Majority rule on j-rich ballot spaces

Abstract

Fishburn (Aggregation and revelation of preferences. North Holland, Amsterdam, pp 201–218, 1979) proved that majority rule on any proper permutation closed $j\hbox {-rich}$ ballot space is the only social choice function satisfying faithfulness, consistency, cancellation, and neutrality. Alós-Ferrer (Soc Choice Welf 27:621–625, 2006) showed that neutrality was not needed for Fishburn’s result as long as the ballot space has no restriction on ballot sizes. In this paper, we show that the Alós-Ferrer result can be extended to a much larger class of ballot spaces.

Appendix

To prove Theorem 4 we will assume that $m \ge 4$, ${\mathcal B}$ is a j-rich ballot space for some $j \ge 3$, and $f: {\mathbb {N}}_0^{{\mathcal B}} \rightarrow P_{ne}(X)$ is a social choice rule satisfying faithfulness, consistency, and cancellation. We want to show that f is majority rule, i.e.,

$$\begin{aligned} f(\pi ) = F_M(\pi ) \end{aligned}$$

for any profile $\pi$. This proof will involve the following notation. For each alternative $x_i\in X$, let the profile $\rho _{x_i}$ be the profile that consists of each of the j sized ballots containing $x_i$. That is,

$$\begin{aligned} \rho _{x_i} = \sum _{B \in {\mathcal B}_j, x_i \in B}B. \end{aligned}$$

For example, if $X = \{x_1, x_2, x_3, x_4\}$ and $j = 3$, then

$$\begin{aligned} \rho _{x_1} = \pi _{\{x_1, x_2, x_3\}} + \pi _{\{x_1, x_2, x_4\}} + \pi _{\{x_1, x_3, x_4\}}. \end{aligned}$$

In this case,

$$\begin{aligned} v(x_1, \rho _{x_1}) = 3\ \text {and}\ v(x_2, \rho _{x_1}) = v(x_3, \rho _{x_1}) = v(x_4, \rho _{x_1}) = 2. \end{aligned}$$

In the general case,

$$\begin{aligned} v(x_i, \rho _{x_i}) = \left( {\begin{array}{c}m - 1\\ j - 1\end{array}}\right) \ \text {and}\ v(x_t, \rho _{x_i}) = \left( {\begin{array}{c}m - 2\\ j - 2\end{array}}\right) \end{aligned}$$

for all $t \ne i$. It follows from consistency and faithfulness that $f(\rho _{x_i}) = \{x_i\}$. For any nonempty subset I of $\{1, \ldots , m\}$ let

$$\begin{aligned} \rho _I = \sum _{i \in I} \rho _{x_i}\ \text {and}\ B_I = \{x_i : i \in I\}. \end{aligned}$$

If $|I| = k$, then

$$\begin{aligned} v(x_i, \rho ) = \left( {\begin{array}{c}m - 1\\ k - 1\end{array}}\right) + (k - 1) \cdot \left( {\begin{array}{c}m - 2\\ k - 2\end{array}}\right) \end{aligned}$$

for all $i \in I$ and

$$\begin{aligned} v(x_t, \rho ) = k \cdot \left( {\begin{array}{c}m - 2\\ k - 2\end{array}}\right) \end{aligned}$$

for all $t \in \{1, \ldots , m\} {\setminus } I$. Therefore,

$$\begin{aligned} F_M(\rho _I) = B_I. \end{aligned}$$

We want to show that $f(\rho _I) = B_I$ as well. The next lemma shows that this equality holds when $|I| = j$ or $j - 1$.

Lemma 1

If $|I| \in \{j, j - 1\}$, then

$$\begin{aligned} f(\rho _I) = B_I. \end{aligned}$$

Proof

Assume $|I| = j$ and note that $B_I \in {\mathcal B}$ since ${\mathcal B}$ is $j\hbox {-rich}$. Let $\alpha$ and $\beta$ be the positive integers satisfying

$$\begin{aligned} \alpha = v(x_i, \rho _I)\ \text {and}\ \beta = v(x_j, \rho _I) \end{aligned}$$

for some $i \in I$ and $j \in \{1, \ldots , m\} {\setminus } I$. Notice that $\alpha > \beta$. Next, let

$$\begin{aligned} \widehat{\rho } = \sum _{i=1}^m \rho _{x_i} \end{aligned}$$

and observe that $f(\widehat{\rho }) = X$ by cancellation. Let $\gamma$ be the positive integer

$$\begin{aligned} \gamma = v(x, \widehat{\rho }) \end{aligned}$$

for any $x \in X$. We now compare the two profiles:

$$\begin{aligned} \gamma \cdot \rho _I\ \quad \ \text {and}\ \quad \ [\beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I]. \end{aligned}$$

For any $x_i \in B_I$,

$$\begin{aligned} v(x_i, \gamma \cdot \rho _I) = \gamma v(x_i, \rho _I) = \gamma \alpha \end{aligned}$$

and

$$\begin{aligned} v(x_i, \beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I) = \beta \gamma + \gamma (\alpha - \beta ) = \gamma \alpha . \end{aligned}$$

Next, for any $x_j \in X {\setminus } B_I$,

$$\begin{aligned} v(x_j, \gamma \cdot \rho _I) = \gamma v(x_j, \rho _I) = \gamma \beta \end{aligned}$$

and

$$\begin{aligned} v(x_j, \beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I) = \beta \gamma . \end{aligned}$$

We now know that

$$\begin{aligned} v(x, \gamma \cdot \rho _I) = v(x, \beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I) \end{aligned}$$

for all $x \in X$. Let $I' = \{1, \ldots , m\} {\setminus } I$ and observe that

$$\begin{aligned} \rho _I + \rho _{I'} = \widehat{\rho }. \end{aligned}$$

Therefore,

$$\begin{aligned} v(x, \gamma \rho _I + \gamma \rho _{I'}) = v(x, \gamma \cdot \widehat{\rho }) = \gamma ^2 \end{aligned}$$

for all $x \in X$. Using the previous equation and fact that

$$\begin{aligned} v(x, \gamma \cdot \rho _I) = v(x, \beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I) \end{aligned}$$

for all $x \in X$ it follows that

$$\begin{aligned} v(x, [ \gamma \rho _{I'} + \beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I)]) = \gamma ^2 \end{aligned}$$

for all $x \in X$ as well. Since f satisfies cancellation we get

$$\begin{aligned} f( \gamma \rho _I + \gamma \rho _{I'}) = f([ \gamma \rho _{I'} + \beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I)]) = X. \end{aligned}$$

Therefore, using consistency (many times) we get

$$\begin{aligned} f(\rho _I)&= f(\gamma \rho _I)\\&=f(\gamma \rho _I + [\gamma \rho _{I'} +\beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I])\\&=f([\gamma \rho _I + \gamma \rho _{I'}] + \beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I)\\&=f(\beta \cdot \widehat{\rho } + \gamma (\alpha - \beta ) \cdot B_I))\\&=f( \gamma (\alpha - \beta ) \cdot B_I))\\&=f(B_I). \end{aligned}$$

Finally, since f is faithful, $f(\rho _I) = f(B_I) = B_I$ and we’re done with the first part of the proof of Lemma 1.

Now assume $|I| = j - 1$ and, as above, let $I' = \{1, \ldots , m\} {\setminus } I$. Using consistency and the first part of this lemma we get

$$\begin{aligned} f\left( \sum _{t \in I'}\ [ \rho _I + \rho _{x_t}] \right) = \bigcap _{t \in I'} \left[ B_I \cup \{x_t\} \right] = B_I. \end{aligned}$$

Note that

$$\begin{aligned} \sum _{t \in I'} [ \rho _I + \rho _{x_t}] = \sum _{i=1}^m \rho _{x_i} + (m - j) \rho _I. \end{aligned}$$

By consistency and cancellation,

$$\begin{aligned} f\left( \sum _{i=1}^m \rho _{x_i} + (m - j) \rho _I \right) = f(\rho _I). \end{aligned}$$

Hence $f(\rho _I) = B_I$.

We are now ready to use Lemma 1 to complete the proof of Theorem 4.

Proof of Theorem 4

Assume that the set

$$\begin{aligned} D = \{ \pi \in {\mathbb {N}}_0^{{\mathcal B}} : f(\pi ) \ne F_M(\pi )\} \end{aligned}$$

is nonempty. So D is the set of profiles where the functions f and $F_M$ disagree. Choose $\rho \in D$ such that $|F_M(\rho )|$ is maximal. This means that if $\pi$ is a profile such that $|F_M(\pi )| > |F_M(\rho )|$, then $f(\pi ) = F_M(\pi )$. Since f is cancellative and $\rho \in D$ it follows that $F_M(\rho ) \ne X$. So

$$\begin{aligned} |F_M(\rho )| \le m - 1. \end{aligned}$$

Assume that there exists $x \in f(\rho )$ such that $x \not \in F_M(\rho )$. We may assume that $x = x_1$. Let

$$\begin{aligned} \ell = \text {max}\ v(\rho ) - v(x_1,\rho ) \end{aligned}$$

and note that $\ell > 0$. Next, let

$$\begin{aligned} \widehat{\rho } = \alpha \rho + \ell \rho _{x_1} \end{aligned}$$

where

$$\begin{aligned} \alpha = \left( {\begin{array}{c}m - 1\\ j - 1\end{array}}\right) - \left( {\begin{array}{c}m - 2\\ j - 2\end{array}}\right) . \end{aligned}$$

Then

$$\begin{aligned} v(x_1, \widehat{\rho }) = \alpha v(x_1, \rho ) + \ell \cdot \left( {\begin{array}{c}m - 1\\ j - 1\end{array}}\right) \end{aligned}$$

and

$$\begin{aligned} v(x_i, \widehat{\rho }) = \alpha v(x_i, \rho ) + \ell \cdot \left( {\begin{array}{c}m - 2\\ j - 2\end{array}}\right) \end{aligned}$$

for all $i \ne 1$. If $v(x_i, \rho ) = \text {max}\ v(\rho ) = [\ell + v(x_1, \rho )]$, then

$$\begin{aligned} v(x_i, \widehat{\rho })&= \alpha \cdot [\ell + v(x_1, \rho )] + \ell \cdot \left( {\begin{array}{c}m - 2\\ j - 2\end{array}}\right) \\&= \alpha v(x_1, \rho ) + \ell \cdot \left( {\begin{array}{c}m - 1\\ j - 1\end{array}}\right) \\&= v(x_1, \widehat{\rho }). \end{aligned}$$

It follows that

$$\begin{aligned} F_M(\widehat{\rho }) = F_M(\rho )\ \cup \{x_1\}. \end{aligned}$$

By our choice of $\rho$ and the fact that $|F_M(\widehat{\rho })| > |F_M(\rho )|$ it follows that

$$\begin{aligned} f(\widehat{\rho }) = F_M(\widehat{\rho }) = F_M(\rho )\ \cup \{x_1\}. \end{aligned}$$

On the other hand, by consistency,

$$\begin{aligned} f(\widehat{\rho }) = f(\rho ) \cap f(\rho _{x_1}) = \{x_1\}. \end{aligned}$$

Since $F_M(\rho )\ \cup \{x_1\} \ne \{x_1\}$ we get a contradiction. It now follows that $f(\rho ) \subset F_M(\rho )$.

Since $f(\rho ) \subset F_M(\rho )$ and $f(\rho ) \ne F_M(\rho )$, there exists $y \in F_M(\rho ) {\setminus } f(\rho )$. Let $x \in X {\setminus } F_M(\rho )$ and $z \in f(\rho )$. We may assume that $x = x_1$, $y = x_2$, and $z = x_3$. As above, let

$$\begin{aligned} \ell = \text {max}\ v(\rho ) - v(x_1,\rho ) \end{aligned}$$

and note that $\ell > 0$. We now introduce the profile

$$\begin{aligned} \mu = \alpha \rho + \ell [ \rho _{x_1} + \rho _{x_2} + \cdots + \rho _{x_j} ]. \end{aligned}$$

By our choice of $\rho$ we know that

$$\begin{aligned} f(\alpha \rho + \ell \rho _{x_1}) = F_M(\alpha \rho + \ell \rho _{x_1}) = F_M(\rho ) \cup \{x_1\}. \end{aligned}$$

Using consistency and Lemma 1,

$$\begin{aligned} f\left( \ell [ \rho _{x_2} + \cdots + \rho _{x_j} ]\right) = \{x_2, \ldots , x_j\}. \end{aligned}$$

Using consistency and the fact that $x_2, x_3 \in F_M(\rho )$ we get

$$\begin{aligned} f(\mu ) = f(\alpha \rho + \ell \rho _{x_1}) \cap f\left( \ell [ \rho _{x_2} + \cdots + \rho _{x_j} ]\right) \supseteq \{x_2,x_3\}. \end{aligned}$$

Next, using consistency and Lemma 1, we get

$$\begin{aligned} f\left( \ell [ \rho _{x_1} + \cdots + \rho _{x_j} ]\right) = \{x_1, \ldots , x_j\}. \end{aligned}$$

Since $x_3 \in f(\rho ) = f(\alpha \rho )$ and $x_3 \in f\left( \ell [ \rho _{x_1} + \cdots + \rho _{x_j} ]\right)$ it follows that

$$\begin{aligned} f(\mu ) = f(\alpha \rho ) \cap f\left( \ell [ \rho _{x_1} + \cdots + \rho _{x_j} ]\right) . \end{aligned}$$

Since $x_2 \not \in f(\rho ) = f(\alpha \rho )$ it follows from the previous equation that $x_2 \not \in f(\mu )$. But this contradicts the fact that $\{x_2, x_3\} \subseteq f(\mu )$. This final contradiction shows that the set $D = \{ \pi \in {\mathbb {N}}_0 : f(\pi ) \ne F_M(\pi )\}$ must be the empty set. Hence $f = F_M$ and we’re done.